**Having a problem when using String Interpolation in the filter pictured. It is telling me Too many positional arguments: 1 allowed, but 2 found
I know what that means, it wants me to remove the String Interpolation that I have in there. I have tried all the bracket combinations I can think of and nothing takes away this error. My String is just a plain string and works fine if I put it directly in the code at this spot but won't work with the interpolation.
Any suggestions on how I can fix this?
Thankyou in Advance For your help **s!
e.ingCatCode == "passedIngCatCode $ingSelSearch"
Use "" quotes and the $ variable inside the quotes.
Related
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I've started work on a Variable banner program, and I've hit a "phantom" syntax error.
name = input('Type here: ')
namelist = list(namelist)
print(namelist)
length_of_name=len(namelist)
asterisk=('*')
for length_of_name:
print (asterisk)
it throws up a syntax error, as I mentioned, but can anyone spot what I did wrong?
The SyntaxError is coming from your for statement. A for statement has to look like for variable in sequence_object:. Here sequence_object is any type that can be iterated, such as a list or tuple, but in your particular case for i in range(0,length_of_name): will make the code syntactically correct. But don't use this because there is a faster way which makes exactly the same output.
Instead of printing a single character in a loop, set asterisk to '*\n' and use print(asterisk*length_of_name, end=''). This prints the same output as the for loop but this multiplies the asterisk character by an integer to make a string that is repeated that many times. That way, you only print once. Setting end to an empty string ensures that a blank line is not printed.
In TI-BASIC, the + operation is overloaded for string concatenation (in this, if nothing else, TI-BASIC joins the rest of the world).
However, any attempt to concatenate involving an empty string raises a Dimension Mismatch error:
"Fizz"+"Buzz"
FizzBuzz
"Fizz"+""
Error
""+"Buzz"
Error
""+""
Error
Why does this occur, and is there an elegant workaround? I've been using a starting space and truncating the string when necessary (doesn't always work well) or using a loop to add characters one at a time (slow).
The best way depends on what you are doing.
If you have a string (in this case, Str1) that you need to concatenate with another (Str2), and you don't know if it is empty, then this is a good general-case solution:
Str2
If length(Str1
Str1+Str2
If you need to loop and add a stuff to the string each time, then this is your best solution:
Before the loop:
" →Str1
In the loop:
Str1+<stuff_that_isn't_an_empty_string>→Str1
After the loop:
sub(Str1,2,length(Str1)-1→Str1
There are other situations, too, and if you have a specific situation, then you should post a simplified version of the relevant code.
Hope this helps!
It is very unfortunate that TI-Basic doesn't support empty strings. If you are starting with an empty string and adding chars, you have to do something like this:
"?
For(I,1,3
Prompt Str1
Ans+Str1
End
sub(Ans,2,length(Ans)-1
Another useful trick is that if you have a string that you are eventually going to evaluate using expr(, you can do "("+Str1+")"→Str1 and then freely do search and replace on the string. This is a necessary workaround since you can't search and replace any text involving the first or last character in a string.
I have a huge string. I need to extract a substring from that that huge string. The conditions are the string starts with either "TECHNICAL" or "JUSTIFY" or "ALIGN" and ends with a number( any number from 1 to 10) followed by period and then followed by space. so for example, I have
string x = "This is a test, again I am testing TECHNICAL: I need to extract this substring starting with testing. 8. This is test again and again and again and again.";
so I need this
TECHNICAL: I need to extract this substring starting with testing.
I was wondering if someone has elegant solution for that.
I was trying to use the regular expression, but I guess I could not figure out the right expresion.
any help will be appreciated.
Thanks in advance.
Try this: #"((?:TECHNICAL|JUSTIFY|ALIGN).*?)(?:[1-9]|10)\. "
I'm using the NVelocity Templating engine to produce a fixed-length field output - you know the kind of thing:
Field Start Pos Field Length Notes
---------- --------- ------------ ---------
Supplier 1 7 Leading Zeros
GRN 8 9 -
...
e.g.
>0001234 123A<
The problem is I'm trying to call String.PadRight() with the overload to specify the leading zero, and NVelocity is having none of it..
This works:
$Document.SupplierCode.PadRight(7)
But this doesn't:
$Document.SupplierCode.PadRight(7,"0")
I've tried:
Single Quotes ('0')
Double Single-Quotes (''0'')
Double Quotes ("0")
Double Double-Quotes (""0"")
Escaping the quotes for all of the above (\"0\")
No Quotes!
All I've found to work from is the NVelocity Homepage, and the Velocity Templating Language Reference page, niether are pointing me at a solution.
Sorry I'm unable to supply or point you somewhere where you can test out your ideas for yourself, but any suggestions you may have will be most welcome!
Thanks for your help ;o)
I'm coping with the same problem at the moment, as far as I understand it is due to the fact that PadLeft and PadRight functions of String class receive the second parameter, the leading "0", as a char, not as a string.
NVelocity allows you to specify the parameter as a string using '0', but in this way internally it generate a cast exception (or something similar), because the parameter is expected as char.
I haven't found yet (I'm just using NVelocity since 1 hour!) a way to specify the parameter as char, at the moment I have just a dirty solution such as applying a Replace(" ", "0") after the PadLeft / PadRight, so the template becomes
$Document.SupplierCode.PadRight(7).Replace(' ', '0')
One solution that a colleague has come up with is to create another property in the Document object that returns the formatted String:
E.g.
Public ReadOnly Property SupplierCodeFormatted() As String
Get
Return Supplier.Code.PadLeft(7, "0")
End Get
End Property